广西南宁市2023届高中毕业班第一次适应性测试数学（理科）试题

南宁市2023届高中毕业班第一次适应性测试数学（理科）参考答案一、选择题：本题共12小题,每小题5分,共60分。在每小题给出的四个选项中,只有一项是符合题目要求的。1.【答案】D1.【解析】因为Axx4,B{xZ3＜x＜7}4,5,6,所以AB5,6,故选D.2.【答案】A3i3i1i24i【解析】由题意z1i3i,可变形为z12i,1i1i1i2则复数z12i,故选A.3.【答案】C3.【解析】依题意C正确.4.【答案】B224.【解析】sin1cos，1cos2cos1,cos2cos20,(cos1)(cos2)0,cos1或cos2(舍)3又sincos1.25.【答案】A1a11【解答】数列满足n整理得常数5.ana1,an1,:1(),3an1an1an11故数列是以3为首项,1为公差的等差数列,所以3(n1)n2,anan所以1n(n1)54故选S55d53125,A.a1226.【答案】C6.【解析】∵X~N1,2,P(0X1)P(1X2)0.287.【答案】Ｄ【解析】已知圆锥的侧面展开图为半径是3的扇形，如图，一只蚂蚁从A点出发绕着圆锥的侧面爬行一圈回到点A的最短距离为AA，设ASA，圆锥底面周长为2，所以2AA32所以，在SAA中，由SASA3,得3AASA2SA22SASAcos13232233()332故选：Ｄ.故P(x0)P(0X1)P(X1)0.280.50.78,故选C.8.【答案】B8.【解析】理数答案第1页共10页π3143143sinsin3sinsincos,sincos,6225225π4πππ2π167sin,cos2cos2cos212sin12,6533662525故选B.9.【答案】Ｃ【解析】由f(x)x2,得f(x)2x,则f(1)2,又f(1)1,所以函数f(x)x2的图象在x1处的切线方程为y12(x1),即y2x1.ex设y2x1与函数g(x)的图象相切于点x,y,a00ex0xgx02,3ea31ee由g(x),可得解得x,ae2,故选Ｃ.x0ae0222gx2x1,0a010【答案】D10【解析】a11,a23,a36,n(n1)12an123nnn22f(n)a1a2a3an1122232n2(123n)21n(n1)(2n1)n(n1)2621n(n1)(n2)611.【答案】A11.【解析】p解法一：如图所示,由题意知直线过抛物线y22px(p0)的焦点F(,0),2p准线方程为x,分别过A,B作准线的垂线,垂足为A,B,2如图,设AAAFt,因为|FB|3|FA|,所以BBBF3t,则BM2t,AB4t,所以ABM60,即直线l的倾斜角等于AFx120,可得直线l的斜率为3,故选A.11FA111解法二：设直线的倾斜角为,由焦比定理,cos3,1BF3112312π依题意cos,,所求直线斜率为k3,故选A.2312.【答案】Bπππππ12.【解析】x[0,π],x,π,令zx,则z,π,333333理数答案第2页共10页1π5π13π13ππ17π由题意sinz在,π上只能有两解z和zπ（）23366,636,*πππ3π因为z,π上必有sinsin2,3322故在上存在满足①成立；(0,π)x1,x2fx1fx22,πz开对应的x（显然在0,π上）一定是最大值点,25π因z对应的x值有可能在0,π上,故②结论错误；2115πππ115解（*）得,所以④成立；当x0,时,z,,由于,6215315362πππππ故z,,,此时ysinz是增函数,从而fx在0,上单调递增,所以31533215③成立,综上,①③④成立,故选B.二、填空题：本题共4小题,每小题5分,共20分。13.【答案】213.【解析】由约束条件作出可行域如图所示,由目标函数z3xy可知,当目标函数过点C(2,4)时,z取得最大值,最大值3242.14.【答案】5详解：取的中点连接如图所示：BCG,AG,D1G,AD1,分别是棱的中点所以E､FAA1､A1D1,EF//AD1,又因为平面平面所以平面EFBEF,AD1BEF,AD1//BEF.因为FD1//BG,FD1=BG,所以四边形为平行四边形所以FBGD1,FB//GD1.又因为平面平面所以平面FBBEF,GD1BEF,GD1//BEF.因为所以平面平面GD1AD1D1,AD1G//BEF.因为点为底面四边形内（包括边界）的一动点直线与PABCD,D1P平面BEF无公共点,22所以P的轨迹为线段AG,则AG215.15.【答案】3PFPFPFsinPFF【解析】由正弦定理得12,所以122sinPF2F1sinPF1F2PF2sinPF1F2即由双曲线的定义可得PF12PF2,PF1PF2PF22a,所以；PF22a,PF14a因为由余弦定理可得222F1PF260,4c16a4a24a2acos60,c2整理可得4c212a2,所以e23,即e3.a2216.【答案】0,1e理数答案第3页共10页x1x【解析】当x0时，fxe2，则fx0，exeexe所以，函数fx在,0上为增函数；当x0时，由y1x20可得y21x2，即y2x21，作出函数fx的图象如下图所示：设过原点且与函数fxx0的图象相切的直线的方程为x，设切点为02，ykxx0,x1ee0x1x所以，切线方程为020，yexx0ex0eex0exx将原点坐标代入切线方程可得020，即e1x0ex0eex0e2x02，eex0ex22xx2构造函数gx，其中x0，则gx0，exeexe2x2所以，函数gx在,0上单调递减，且gee，exex21x由02，解得，所以，0，gx0ex0ekxee1ex0ee0而函数fx1x2x0的渐近线方程为yx，设直线yx与ye1x的夹角为，设直线ye1x的倾斜角为，3πtantan3π41e12则tantan1，3π41tantan1e1e422结合图形可知，0tanMON1.故答案为：0,1.ee三、解答题：共70分。解答应写出文字说明、证明过程或演算步骤。第17～21题为必考题,每个试题考生都必须作答。第22、23题为选考题,考生依据要求作答。π【答案】（）B（）(5,6]17.132【解析】（1）由(bc)(sinBsinC)(sinAsinC)a根据正弦定理可得(bc)(bc)(ac)a,·························································1分所以,a2c2b2ac,··················································································2分理数答案第4页共10页a2c2b21由余弦定理可得cosB,··························································3分2ac2∵B(0,π),································································································4分π因此B．·······························································································5分3(2)由余弦定理,得b2a2c22accosB,3a2c2ac,····································6分即a2c23acacb3由正弦定理得2,sinAsinCsinB3,·····················································7分22即a2sinA,c2sinC,又CA,所以322ac4sinAsinC4sinAsinA23sinAcosA2sinA33sin2Acos2A12sin2A1.······················································9分60A2由,解得A,································································10分2620A32所以5所以1分2A,sin2A,1,··············································1166662所以ac(2,3],所以a2c23ac(5,6].·······················································12分18.(І)证明：∵MN分别是AD、AC的中点∴MN//DC，又∵DC//EB∴MN//EB，∴E、M、N、B四点共面.·····················································3分在图1中,由AB6,AE2EB得AE4,BE2,∵AB//DC,ABC90,BCDC2.∴四边形DEBC是正方形，∴DEAB.在图2中，平面AED平面DEBC，平面AED平面DEBCED,AEED.∴AE平面DEBC，∴AEDC又∵DCDE,DEAEE,∴DC平面AED，∴平面ADC平面AED.··················································6分(ІІ)由图易知直线EA、EB、ED两两垂直,以E为原点，分别以EB、ED、EA所在直线为x轴、y轴、z轴,建立空间直角坐标系Exyz，则B(2,0,0),C(2,2,0),D(0,2,0),理数答案第5页共10页设G(0,0,t),(0t4).得CG(2,2,t),BD(2,2,0),BG(2,0,t)·····················8分设n(x,y,z)是平面BDG的